c) \(M=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{1}{2}.\frac{4}{4}.\frac{6}{6}...\frac{100}{100}=\frac{1}{2}\)

a) M . N = \(\left(\frac{1}{2.}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)=\frac{1.2.3.4....100}{2.3.4.5...101}=\frac{1}{101}\)

C=1/2*2+1/4*4+1/6*6+...+1/100*100.

C<1/4+1/2*4+1/4*6+1/6*8+...+1/98*100.

C<1/4+1/2*(2/2*4+2/4*6+2/6*8+...+2/98*100).

C<1/4+1/2*(1/2-1/4+1/4-1/6+1/6-1/8+...+1/98-1/100).

C<1/4+1/2*(1/2-1/100).

C<1/4+1/2*49/100.

C<1/4+49/200.

C<1/4+50/200=1/2.

Vậy C<1/2.

ta có \(\frac{1}{2\cdot2}+\frac{1}{4\cdot4}+\frac{1}{6\cdot6}+.........+\frac{1}{100\cdot100}\)

\(< \frac{1}{4}+\frac{1}{2x4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+........+\frac{1}{98\cdot100}\)

\(\frac{1}{4}+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{98\cdot100}\right)\)

=\(\frac{1}{4}+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{4}+\frac{1}{2}\cdot\frac{49}{100}=\frac{1}{4}+\frac{49}{200}\)

tự làm nốt

Ta có: \(\frac{1}{5^2}< \frac{1}{4.5};\frac{1}{6^2}< \frac{1}{5.6};...;\frac{1}{100^2}< \frac{1}{99.100}\)

Cộng vế với vế ta được: \(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}< \frac{6}{24}=\frac{1}{4}\)(1)

Tương tự: \(\frac{1}{5^2}>\frac{1}{5.6};\frac{1}{6^2}>\frac{1}{6.7};...;\frac{1}{100^2}>\frac{1}{100.101}\)

Cộng vế với vế ta được \(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{96}{576}=\frac{1}{6}\)(2)

Từ (1) và (2) =>đpcm

Đặt \(A=\displaystyle\sum_{i=5}^{100}\frac{1}{i^2}\)

\(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

\(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}>\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\)

sao cái mã latex ko hiển thị nhờ :(( A là cái biểu thức ở giữa nhé